Came up with the exact same solution. Thanks for the warning message as well.

I am wondering what kind of data structure of count[] using in the code? And thanks for the warning, it does make sense.

can anyone explain me last print statement...thanks in advance

What a brilliant solution. I love it.

I did it similarly to you at first but then I tried to optimize it from O(2n) (for loop and list.index()) to O(n) and came up with this:

def migratoryBirds(n, ar):
    ids = [0] * 5
    maxN = 0
    id = 1
    for i in range(n):
        ids[ar[i] - 1] += 1
        if(ids[ar[i] - 1] > maxN):
            maxN = ids[ar[i] - 1]
            id = ar[i]
        elif(ids[ar[i] - 1] == maxN and ar[i] < id):
            id = ar[i]
    return id

n = int(input().strip())
ar = list(map(int, input().strip().split(' ')))
a = ar.count(1)
b = ar.count(2)
c = ar.count(3)
d = ar.count(4)
e = ar.count(5)
li=[a,b,c,d,e]
print(1+li.index(max(li)))

what is the Time Complexity of this code?

This, to me, is what places like this should be all about - clear code which works, but also which points out features that others (like me) may not know about - the index method is really useful, and I'd not seen it before. My code did something similar, but used more code to do the same thing. Thanks for posting it and the comment.

Nice [0]*6 syntax! My algorithm is the same as yours, but not written in a Pythonic way...

for those who are beginners, I barely know what log(n) is here I brought really simple answer, I passed all tests too. Feel free to ask me questions, but i guess you guyz already know

public static void main(String args[]){
		Scanner scan = new Scanner(System.in);
		int n = scan.nextInt();
		int[] arr = new int[5];
		for(int i = 0;i<n;i++){
			switch (scan.nextInt()){
			case 1 :
				arr[0]++;
				break;
			case 2 :
				arr[1]++;
				break;
			case 3 :
				arr[2]++;
				break;
			case 4 :
				arr[3]++;
				break;
			case 5 :
				arr[4]++;
				break;
			}
		}
		int highest = 0;
		int answer = 0;
		for(int i = 0;i<arr.length;i++){
			if(arr[i]>highest){
				highest = arr[i];
				answer= i+1;
			}
		}
		System.out.println(answer);
	}

Please let me know what you think of my code in C#.

using System;
using System.Collections.Generic;
using System.IO;
using System.Linq;

public class Solution {
    #region Private Property
    /// <summary>
    /// Gets or sets the Length.
    /// </summary>
    private static int Length { get; set; }
    #endregion

    #region Methods
    /// <summary>
    /// Returns the Max count of birds.
    /// </summary>
    /// <param name='arr'></param>
    private static String migratoryBirds(String[] arr) {
        // Complete this function
        return arr.GroupBy(a => a).Select(b => new{ Bird = b.Key, Count = b.Count() }).OrderByDescending(c => c.Count).ThenBy(d => d.Bird).FirstOrDefault().Bird;
    }
    #endregion

    public static void Main(String[] args) {
        Length = Convert.ToInt32(Console.ReadLine());
        Console.WriteLine(migratoryBirds(Console.ReadLine().Split(' ')));
    }
}

Pretty cool!

how do you think max function works?

Question. Why is the list 'count' of length '6'? We only have 5 classes of bird types right? Also what does max(count) return when all classes of birds have equal occurences? Does it return the one with the smallest ID?

Your code is doing more than this one:

input()
count = [0]*6
for t in map(int,input().strip().split()):
    count[t] += 1
maxIndex = 0
for i,v in enumerate(count):
   if v>count[maxIndex]:
   	maxIndex = i
print(maxIndex)

The reason is that your code is looking for the maximum the looking again to retrieve the index: count.index(max(count))

The code I wrote skip one loop by working directly with the index.

It shows also that when you want efficience, you may have to sacrifice legibility.

So, for the given problem, your solution ace it. For a more general situation, one has to be warry that you implement something very specific to the problem at hand:

• Fixed number of bird Ids that are consecutive
• Ids are numeric

Would break if "names" where given instead of Ids or if ther was a lot of Ids.

The code you provided is well suited for small diversity of bird tagged by numerical ids.

I have another solution, I think It is also correct and eazy.

def migratoryBirds(n, ar):
    m = [4,3,2,1,0]
    for b in ar:
        m[b-1] += 10
    return 5 - max(m) % 10

don't you get tired of writing" elegant code, concise code, optimal code blah blah and blah". just put the damn code, explain it and that's it.. story over!!

Why do you need the map? What is wrong with:

def build_bird_freq(arr: list):
    bird_freq = [0, 0, 0, 0, 0, 0]
    for i in range(len(arr)):
        bird_freq[arr[i]] += 1
    return bird_freq.index(max(bird_freq))

Here is my O(n) inplace solution:

static int migratoryBirds(int[] ar) {

for(int i=0;i<ar.length;i++)
   ar[(ar[i]%10)-1] = ar[(ar[i]%10)-1]+10;
int max = -1,q=0;
for(int i=0;i<ar.length;i++){
    if(max<ar[i]/10){
        max = ar[i]/10;
        q=i+1;
    }
}
     return q;
}

Nice idea Sir ! I can say, you are actually encoding the array & then decoding it back.. great thought !

Even simpler and perhaps faster for a huge list:

return max(set(ar), key=ar.count)

This is so ingenious. As a Python beginner I am amazed to see how simple and elegant your solution is. Awesome stuff lol

max() is also a linear function, so your soluion will also traverses the list twice.

Thank you :). Sometimes, the simplest way is the best way, both in code and in life.

The "count" list must actually be of length max(given_list_elements)+1,as the index in python starts from '0'.The zeroth index value of count is never edited if the bird IDs start from '1'.However,that's not gonna affect the output.So it's ok : )

I didnt think of creating a counter list. Thank you for teaching me a new way

Can anyone explain , why he used 6 here.

count = [0]*6 It may fail when the input list contains value more than 6

eg: [10,10,1,2,3,4]

Thanks for the warnming man. We need friends like you more!

great!

can you please explain in detail what the first for loop does?

ok I found out the first for loop gives us the count of the respective elements in another array Example: ar[0] = 2; ar[1] = 4; ar[2] = 3; ar[3] = 2; ar[4] = 3; ar[5] = 1; ar[6] = 2; ar[7] = 1; ar[8] = 3; ar[9] = 3;

    ary[0] = 1;
    ary[1] = 1;
    ary[2] = 1;
    ary[3] = 2;
    ary[4] = 2;
    ary[5] = 1;
    ary[6] = 3;
    ary[7] = 2;
    ary[8] = 3;
    ary[9] = 4;

    but my doubt now is how did you come up with this logic about the first for loop??????

Is there any advantage of:

• Replacing 'n' space allocation in ary[] by '6' ?

• Replacing 'n' in second for loop by '6' ?

as it is mentioned only '5' identity no's .

int ary[] = new int[n]; can be replaced by

int ary[] = new int[5];
or int ary[] = new int[6];

to have the same indeces

Thanks man

Nice solution. A little optimization.

static int migratoryBirds(int n, int[] ar) {
        // Complete this function
        
        int []arr = new int [6];
        
        for(int id:ar){
            arr[id]++;
        }
        
        int maxValue = 0; 
        int maxPos=0;
    
        for(int i =1;i<6;i++){
            if(arr[i]>maxValue){
                maxValue = arr[i];
                maxPos =  i;
            }       
        }
        return maxPos;
 }

What is the problem with my solution?

static int migratoryBirds(int n, int[] ar) {
        int zero=0;
        int[] array=new int[5];
        for(int i : ar){
            array[i-1]++;
        }
        for(int i:array){
            if(i==0) zero++;
        }
        
        return 5-zero;
    }

what does arr[ar[i]]++ do?

sir i aslo did the same as you but i want to ask some thing there is one condition in question If two or more types of birds are equally common, choose the type with the smallest ID number.

according to this condition if the type 2 will repeat 3 time and the type 4 will aslo repeat 3 time than the smallest index value is 2 but the solution of your and my always return the index 4

can you please see it once .....??

Hey I have a doubt.Can you tell me where you are checking the condition if two birds have equal count then printing the bird smaller index count

And also your output will fail if u enter the input as

4 then 5 5 6 6

salute , maja aaaya.

Your ary[] array can be of the size 6 instead of n

You dont need a max variable. You can compare ary[i] with ary[maxpos].

Hey brother, Could you please explain how's ary[ar[i]]++; does work; i don't get it. Because your answer of ary[ar[i]]++; 0 1 0 1 3 1

Hey brother, Could you please explain how's ary[ar[i]]++; does work; i don't get it. Because your answer of ary[ar[i]]++; 0 1 0 1 3 1

thank you sir

Does this work with the last test case?

Did it similarly, but didn't pass all tests :P

Where is the hurry? Why the oneline stunt? Hahha! You're a pro!

This... shouldn't work.

set() is a hashset and has no fixed order. it should be sorted() instead of list()

print(sorted(reversed(sorted(set(types))), key=types.count)[-1])

Because there are only 5 types of birds, it can just be

print(sorted(list("54321"), key=types.count)[-1])

Here's my solution:

input()
types=input().split()
print(-max((types.count(i),-int(i)) for i in set(types))[1])

It's worth adding that this is sub-optimal. The Counter constructor will iterate the list, then most_common will sort it in reverse order using a heap.

Solved it like this as well. However, documentation says:

most_common([n]) Return a list of the n most common elements and their counts from the most common to the least. If n is omitted or None, most_common() returns all elements in the counter. Elements with equal counts are ordered arbitrarily

While most_common(1) gave me the corret id, the above seems to indicate that this doesnt have to be true.

could you explain

for(int i = 0; i<n; i++)
        {
            id[types[i]-1]++;
        }

is what meaning?

nice approach dude...

With an extra variable you can keep track of the position with the highest count, and remove the second loop.

Unneded complexity: you had better only memorize the frequencies and remove the types pointer. And not freeing dynamically allocated memory is a bad practice.

code is nice

Awesome

please explain me ur code

C++ had the same issue.

:-):-):-)

result.append(i)

Can you explain what the last line is doing? My linq is pretty rusty. Specifically:

Q => Q.Value == dic.Values.Max()).Key